Question

rationalise the denominator
root2÷(root2 + root3 - root5)
​

Answer 1

Here is the answer.

Hope it helps!!

Answer 2

The given question is that we have to rationalize the given expression.

The given expression is

$\frac{ \sqrt{2} }{ \sqrt{2} + \sqrt{3} - \sqrt{5} }$

on rationalising the denominator, we have to multiply the entire expression with the conjugate term of denominator.

$\frac{ \sqrt{2} }{ (\sqrt{2} + \sqrt{3} )- \sqrt{5} } \\ \frac{ \sqrt{2} }{( \sqrt{2} + \sqrt{3} - \sqrt{5} } \times \frac{ \sqrt{2} }{( \sqrt{2} + \sqrt{3} ) + \sqrt{5} } \\ \frac{ \sqrt{2} ( \sqrt{2 } + \sqrt{3}) + \sqrt{5} }{ {( \sqrt{2} + \sqrt{3}) }^{2} - { \sqrt{5} }^{2} } \\ \frac{ { \sqrt{2} }^{2} + \sqrt{6} + \sqrt{10} }{ { \sqrt{2} }^{2} + { \sqrt{3} }^{2} + 2 \times \sqrt{6} - 5 } \\ \frac{2 + \sqrt{6} + \sqrt{10} }{5 + 2 \sqrt{6} - 5 } \\ \frac{2 + \sqrt{6} + \sqrt{10} }{2 \sqrt{6} }$

On rationalising we get the above factor.

again on rationalising the factor obtained now will be

$\frac{2 + \sqrt{6} + \sqrt{10} }{2 \sqrt{6} } \times \frac{2 \sqrt{6} }{2 \sqrt{6} } \\ \frac{2 \sqrt{6} (2 + \sqrt{6} + \sqrt{10} )}{ {(2 \sqrt{6} )}^{2} } \\ \frac{4 \sqrt{6} + 2 \times { \sqrt{6} }^{2} + 2 \sqrt{60} }{4 \times 6} \\ \frac{4 \sqrt{6} + 12 + 2 \sqrt{60} }{24} \\ \frac{4\sqrt{6} + 12+ 2 \sqrt{4 + 15} }{24} \\ \frac{4 \sqrt{6} + 12 + 2 \times 2 \times \sqrt{15} }{24} \\ \frac{4( \sqrt{6} + 3 + \sqrt{15} }{24} \\ \frac{ \sqrt{6} + 3 + \sqrt{15}) }{6}$

Therefore, the final answer obtained is

$\frac{3 + \sqrt{6} + \sqrt{15} }{6}$

# spj2

we can find the similar questions through the link given below

https://brainly.in/question/16843569?referrer=searchResults